Question

A Particle is Found to be in a stute 2/ 1 q)pind the hormdligation Constan p is the Pradbalbiliter thut amensurement o Land Lz will give 6K and h festectivels alalate

Answer 1

(a) Normalisation condition

$\int_0^{2\pi}\int_0^\pi\psi^*\psi\sin\theta\, d\theta d\theta d\phi=1$

Using the fact that Y's are orthonormal, we get

$|N|^2\left ( \frac{1}{3}+\frac{2}{3}+4 \right )=1$

$N=\frac{1}{\sqrt{5}}$

(b)

$L^2=\ell(\ell+1)\hbar^2$

which means

$\ell=2$

and

$L_z=m\hbar$

$\Rightarrow m=1$

Check if the state has a term of Y_2,1

Thus, the probability to be in this state is

$|N|^2\cdot(\textup{coeff. of } Y_{2,1})^2=\frac{4}{5}$

(c)

$L_-L_+=(L_x-iL_y)(L_x+iL_y)=L_x^2+L_y^2+i[L_x,L_y]=L^2-L_z^2-\hbar L_z$

$L^2|\psi\rangle=N\left (\sqrt\frac{1}{3}\cdot2\hbar^2\,Y_{1,0}+\sqrt\frac{2}{3}\cdot12\hbar^2\,Y_{3,1}-2\cdot6\hbar^2\,Y_{2,1} \right )$

$\left (L_z^2+\hbar L_z \right )|\psi\rangle=N\left (\sqrt\frac{1}{3}\cdot0\hbar^2\,Y_{1,0}+\sqrt\frac{2}{3}\cdot2\hbar^2\,Y_{3,1}-2\cdot2\hbar^2\,Y_{2,1} \right )$

$L_-L_+|\psi\rangle=N\hbar^2\left (\sqrt\frac{1}{3}\cdot2\,Y_{1,0}+\sqrt\frac{2}{3}\cdot10\,Y_{3,1}-2\cdot4\,Y_{2,1} \right )$

$\langle\psi|L_-L_+|\psi\rangle=N^2\hbar^2\left (\frac{1}{3}\cdot2+\frac{2}{3}\cdot10+4\cdot4 \right )$

$\langle\psi|L_-L_+|\psi\rangle=\frac{14}{3}\hbar^2$